Lessons in Electric Circuits Bipolar Junction Transistor

7/25/2019 Lessons in Electric Circuits Bipolar Junction Transistor

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Lessons In Electric Circuits -- Volume III

Chapter 4

BIPOLAR JUNCTION TRANITOR

!!! INCO"PLETE !!!

Intro#uction

The invention of the bipolar transistor in 1948 ushered in a revolution in

electronics. Technical feats previously requiring relatively large, mechanicallyfragile, power-hungry vacuum tubes were suddenly achievable with tiny,

mechanically rugged, power-thrifty specs of crystalline silicon. This revolutionmade possible the design and manufacture of lightweight, ine!pensive electronic

devices that we now tae for granted. "nderstanding how transistors function is ofparamount importance to anyone interested in understanding modern electronics.

#y intent here is to focus as e!clusively as possible on the practical function andapplication of bipolar transistors, rather than to e!plore the quantum world of

semiconductor theory. $iscussions of holes and electrons are better left to anotherchapter in my opinion. %ere & want to e!plore how to usethese components, not

analy'e their intimate internal details. & don(t mean to downplay the importance ofunderstanding semiconductor physics, but sometimes an intense focus on solid-

state physics detracts from understanding these devices( functions on a componentlevel. &n taing this approach, however, & assume that the reader possesses a certain

minimum nowledge of semiconductors) the difference between *+* and ** doped

semiconductors, the functional characteristics of a + diode /unction, and themeanings of the terms *reverse biased* and *forward biased.* &f these concepts areunclear to you, it is best to refer to earlier chapters in this boo before proceeding

with this one.

0 bipolar transistor consists of a three-layer *sandwich* of doped e!trinsic

semiconductor materials, either +--+ or -+-. ach layer forming the transistorhas a specific name, and each layer is provided with a wire contact for connection

to a circuit. 2hown here are schematic symbols and physical diagrams of these two

transistor types)


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ipolar transistors are so named because the controlled current must go

through twotypes of semiconductor material) + and . The current consistsof both electron and hole flow, in different parts of the transistor.

ipolar transistors consist of either a +--+ or an -+- semiconductor*sandwich* structure.

The three leads of a bipolar transistor are called theEmitter,Base,

and Collector.

Transistors function as current regulators by allowing a small current

to controla larger current. The amount of current allowed between collector

and emitter is primarily determined by the amount of current movingbetween base and emitter.

&n order for a transistor to properly function as a current regulator, the

controlling base current and the controlled collector currents must be

going in the proper directions) meshing additively at the emitter andgoing againstthe emitter arrow symbol.

The transistor as a s&itch

ecause a transistor(s collector current is proportionally limited by its base current,it can be used as a sort of current-controlled switch. 0 relatively small flow of

electrons sent through the base of the transistor has the ability to e!ert control overa much larger flow of electrons through the collector.

2uppose we had a lamp that we wanted to turn on and off by means of a switch.

2uch a circuit would be e!tremely simple)

3or the sae of illustration, let(s insert a transistor in place of the switch to show

how it can control the flow of electrons through the lamp. emember that thecontrolled current through a transistor must go between collector and emitter. 2ince

it(s the current through the lamp that we want to control, we must position thecollector and emitter of our transistor where the two contacts of the switch are now.


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e must also mae sure that the lamp(s current will move againstthe direction of

the emitter arrow symbol to ensure that the transistor(s /unction bias will be correct)

&n this e!ample & happened to choose an + transistor. 0 ++ transistor could

also have been chosen for the /ob, and its application would loo lie this)

The choice between + and ++ is really arbitrary. 0ll that matters is that theproper current directions are maintained for the sae of correct /unction biasing

electron flow going againstthe transistor symbol(s arrow.

:oing bac to the + transistor in our e!ample circuit, we are faced with the need

to add something more so that we can have base current. ithout a connection tothe base wire of the transistor, base current will be 'ero, and the transistor cannot

turn on, resulting in a lamp that is always off. emember that for an + transistor,base current must consist of electrons flowing from emitter to base against the

emitter arrow symbol, /ust lie the lamp current. +erhaps the simplest thing to dowould be to connect a switch between the base and collector wires of the transistor

lie this)

&f the switch is open, the base wire of the transistor will be left *floating* not

connected to anything and there will be no current through it. &n this state, the

transistor is said to be cutoff. &f the switch is closed, however, electrons will be ableto flow from the emitter through to the base of the transistor, through the switch


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ven a microphone of sufficient voltage and current output could be used to turn

the transistor on, provided its output is rectified from 05 to $5 so that the emitter-

base + /unction within the transistor will always be forward-biased)

The point should be quite apparent by now) anysufficient source of $5 current

may be used to turn the transistor on, and that source of current need only be afraction of the amount of current needed to energi'e the lamp. %ere we see the

transistor functioning not only as a switch, but as a true amplifier) using a relativelylow-power signal to controla relatively large amount of power. +lease note that the

actual power for lighting up the lamp comes from the battery to the right of theschematic. &t is not as though the small signal current from the solar cell,

thermocouple, or microphone is being magically transformed into a greater amountof power. ather, those small power sources are simply controllingthe battery(s

power to light up the lamp.

REVIE$%

Transistors may be used as switching elements to control $5 power to a

load. The switched controlled current goes between emitter and collector,

while the controlling current goes between emitter and base.

hen a transistor has 'ero current through it, it is said to be in a stateof cutofffully nonconducting.


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hen a transistor has ma!imum current through it, it is said to be in a state

ofsaturationfully conducting.

"eter chec' o( a transistor

ipolar transistors are constructed of a three-layer semiconductor *sandwich,*

either ++ or +. 0s such, they register as two diodes connected bac-to-bacwhen tested with a multimeter(s *resistance* or *diode chec* functions)

%ere &(m assuming the use of a multimeter with only a single continuity rangeresistance function to chec the + /unctions. 2ome multimeters are equipped

with two separate continuity chec functions) resistance and *diode chec,* eachwith its own purpose. &f your meter has a designated *diode chec* function, use

that rather than the *resistance* range, and the meter will display the actual forwardvoltage of the + /unction and not /ust whether or not it conducts current.


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#eter readings will be e!actly opposite, of course, for an + transistor, with both

+ /unctions facing the other way. &f a multimeter with a *diode chec* function isused in this test, it will be found that the emitter-base /unction possesses a slightly

greater forward voltage drop than the collector-base /unction. This forward voltagedifference is due to the disparity in doping concentration between the emitter andcollector regions of the transistor) the emitter is a much more heavily doped piece

of semiconductor material than the collector, causing its /unction with the base toproduce a higher forward voltage drop.

6nowing this, it becomes possible to determine which wire is which on an

unmared transistor. This is important because transistor pacaging, unfortunately,is not standardi'ed. 0ll bipolar transistors have three wires, of course, but the

positions of the three wires on the actual physical pacage are not arranged in anyuniversal, standardi'ed order.

2uppose a technician finds a bipolar transistor and proceeds to measure continuitywith a multimeter set in the *diode chec* mode. #easuring between pairs of wires

and recording the values displayed by the meter, the technician obtains thefollowing data)


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#eter touching wire 1 = and > -) *;7*

#eter touching wire 1 - and > =) *;7*

#eter touching wire 1 = and ? -) @.ABB volts

#eter touching wire 1 - and ? =) *;7*

#eter touching wire > = and ? -) @.A>1 volts

#eter touching wire > - and ? =) *;7*

The only combinations of test points giving conducting meter readings are wires 1and ? red test lead on 1 and blac test lead on ?, and wires > and ? red test lead

on > and blac test lead on ?. These two readings mustindicate forward biasing ofthe emitter-to-base /unction @.ABB volts and the collector-to-base /unction @.A>1

volts.

ow we loo for the one wire common to both sets of conductive readings. &t must

be the base connection of the transistor, because the base is the only layer of the

three-layer device common to both sets of + /unctions emitter-base and collector-base. &n this e!ample, that wire is number ?, being common to both the 1-? and the>-? test point combinations. &n both those sets of meter readings, the black- meter

test lead was touching wire ?, which tells us that the base of this transistor is madeof -type semiconductor material blac C negative. Thus, the transistor is an ++

type with base on wire ?, emitter on wire 1 and collector on wire >)


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+lease note that the base wire in this e!ample is notthe middle lead of thetransistor, as one might e!pect from the three-layer *sandwich* model of a bipolar

transistor. This is quite often the case, and tends to confuse new students ofelectronics. The only way to be sure which lead is which is by a meter chec, or by

referencing the manufacturer(s *data sheet* documentation on that particular partnumber of transistor.

6nowing that a bipolar transistor behaves as two bac-to-bac diodes when tested

with a conductivity meter is helpful for identifying an unnown transistor purely bymeter readings. &t is also helpful for a quic functional chec of the transistor. &f the

technician were to measure continuity in any more than two or any less than two ofthe si! test lead combinations, he or she would immediately now that the transistorwas defective or else that it wasn'ta bipolar transistor but rather something else --

a distinct possibility if no part numbers can be referenced for sure identificationD.%owever, the *two diode* model of the transistor fails to e!plain how or why it acts

as an amplifying device.

To better illustrate this parado!, let(s e!amine one of the transistor switch circuitsusing the physical diagram rather than the schematic symbol to represent the

transistor. This way the two + /unctions will be easier to see)


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0 grey-colored diagonal arrow shows the direction of electron flow through the

emitter-base /unction. This part maes sense, since the electrons are flowing fromthe -type emitter to the +-type base) the /unction is obviously forward-biased.

%owever, the base-collector /unction is another matter entirely. otice how thegrey-colored thic arrow is pointing in the direction of electron flow upwards

from base to collector. ith the base made of +-type material and the collector of-type material, this direction of electron flow is clearly bacwards to the direction

normally associated with a + /unctionD 0 normal + /unction wouldn(t permit this*bacward* direction of flow, at least not without offering significant opposition.

%owever, when the transistor is saturated, there is very little opposition to electrons

all the way from emitter to collector, as evidenced by the lamp(s illuminationD

5learly then, something is going on here that defies the simple *two-diode*e!planatory model of the bipolar transistor. hen & was first learning about

transistor operation, & tried to construct my own transistor from two bac-to-bacdiodes, lie this)


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#y circuit didn(t wor, and & was mystified. %owever useful the *two diode*

description of a transistor might be for testing purposes, it doesn(t e!plain how atransistor can behave as a controlled switch.

hat happens in a transistor is this) the reverse bias of the base-collector /unctionprevents collector current when the transistor is in cutoff mode that is, when there

is no base current. %owever, when the base-emitter /unction is forward biased bythe controlling signal, the normally-blocing action of the base-collector /unction is

overridden and current is permitted through the collector, despite the fact thatelectrons are going the *wrong way* through that + /unction. This action is

dependent on the quantum physics of semiconductor /unctions, and can only taeplace when the two /unctions are properly spaced and the doping concentrations of

the three layers are properly proportioned. Two diodes wired in series fail to meetthese criteria, and so the top diode can never *turn on* when it is reversed biased,no matter how much current goes through the bottom diode in the base wire loop.

That doping concentrations play a crucial part in the special abilities of the

transistor is further evidenced by the fact that collector and emitter are notinterchangeable. &f the transistor is merely viewed as two bac-to-bac +

/unctions, or merely as a plain -+- or +--+ sandwich of materials, it may seemas though either end of the transistor could serve as collector or emitter. This,

however, is not true. &f connected *bacwards* in a circuit, a base-collector currentwill fail to control current between collector and emitter. $espite the fact that both

the emitter and collector layers of a bipolar transistor are of the samedoping typeeither or +, they are definitely not identicalD

2o, current through the emitter-base /unction allows current through the reverse-

biased base-collector /unction. The action of base current can be thought of as*opening a gate* for current through the collector. #ore specifically, any given


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amount of emitter-to-base currentpermits a limited amountof base-to-collector

current. 3or every electron that passes through the emitter-base /unction and onthrough the base wire, there is allowed a certain, restricted number of electrons to

pass through the base-collector /unction and no more.

&n the ne!t section, this current-limiting behavior of the transistor will be

investigated in more detail.

REVIE$%

Tested with a multimeter in the *resistance* or *diode chec* modes, a

transistor behaves lie two bac-to-bac + diode /unctions.

The emitter-base + /unction has a slightly greater forward voltage drop

than the collector-base + /unction, due to more concentrated doping of theemitter semiconductor layer.

The reverse-biased base-collector /unction normally blocs any current from

going through the transistor between emitter and collector. %owever, that/unction begins to conduct if current is drawn through the base wire. ase

current can be thought of as *opening a gate* for a certain, limited amount ofcurrent through the collector.

Acti)e mo#e operation

hen a transistor is in the fully-off state lie an open switch, it is said to

be cutoff. 5onversely, when it is fully conductive between emitter and collectorpassing as much current through the collector as the collector power supply and

load will allow, it is said to besaturated. These are the two modes of operatione!plored thus far in using the transistor as a switch.

%owever, bipolar transistors don(t have to be restricted to these two e!treme modesof operation. 0s we learned in the previous section, base current *opens a gate* for

a limited amount of current through the collector. &f this limit for the controlledcurrent is greater than 'ero but less than the ma!imum allowed by the power supply

and load circuit, the transistor will *throttle* the collector current in a modesomewhere between cutoff and saturation. This mode of operation is called

the activemode.

0n automotive analogy for transistor operation is as follows) cutoffis the condition

where there is no motive force generated by the mechanical parts of the car to mae

it move. &n cutoff mode, the brae is engaged 'ero base current, preventingmotion collector current.Active modeis when the automobile is cruising at a


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constant, controlled speed constant, controlled collector current as dictated by the

driver. Saturationis when the automobile is driving up a steep hill that prevents itfrom going as fast as the driver would wish. &n other words, a *saturated*

automobile is one where the accelerator pedal is pushed all the way down basecurrent calling for more collector current than can be provided by the power

supplyEload circuit.

&(ll set up a circuit for 2+&5 simulation to demonstrate what happens when a

transistor is in its active mode of operation)

*F* is the standard letter designation for a transistor in a schematic diagram, /ust as** is for resistor and *5* is for capacitor. &n this circuit, we have an +

transistor powered by a battery G1 and controlled by current through a currentsource&1. 0 current source is a device that outputs a specific amount of current,

generating as much or as little voltage as necessary across its terminals to ensurethat e!act amount of current through it. 5urrent sources are notoriously difficult to

find in nature unlie voltage sources, which by contrast attempt to maintain aconstant voltage, outputting as much or as little current in the fulfillment of that

tas, but can be simulated with a small collection of electronic components. 0s we

are about to see, transistors themselves tend to mimic the constant-current behaviorof a current source in their ability to regulatecurrent at a fi!ed value.

&n the 2+&5 simulation, &(ll set the current source at a constant value of >@ H0,

then vary the voltage source G1 over a range of @ to > volts and monitor howmuch current goes through it. The *dummy* battery Gammeter with its output of @

volts serves merely to provide 2+&5 with a circuit element for currentmeasurement.


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bipolar transistor simulation

i1 0 1 dc 20u

q1 2 1 0 mod1

vammeter 3 2 dc 0

v1 3 0 dc

.model mod1 npn

.dc v1 0 2 0.05.plot dc i(vammeter)

.end

type npn

is 1.00E-16

bf 100.000

nf 1.000

br 1.000

nr 1.000

v1 i(ammeter) -1.000E-03 0.000E00 1.000E-03 2.000E-03

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

0.000E00 -1.!"0E-05 . # . .

5.000E-02 !.1""E-05 . .# . .

1.000E-01 6.1!5E-0$ . . # . .

1.500E-01 1.526E-03 . . . # .

2.000E-01 1.!1$E-03 . . . #.

2.500E-01 1.!"%E-03 . . . #

3.000E-01 1.!!"E-03 . . . #

3.500E-01 2.000E-03 . . . #

$.000E-01 2.000E-03 . . . #$.500E-01 2.000E-03 . . . #

5.000E-01 2.000E-03 . . . #

5.500E-01 2.000E-03 . . . #

6.000E-01 2.000E-03 . . . #

6.500E-01 2.000E-03 . . . #

%.000E-01 2.000E-03 . . . #

%.500E-01 2.000E-03 . . . #

".000E-01 2.000E-03 . . . #

".500E-01 2.000E-03 . . . #

!.000E-01 2.000E-03 . . . #

!.500E-01 2.000E-03 . . . #

1.000E00 2.000E-03 . . . #

1.050E00 2.000E-03 . . . #1.100E00 2.000E-03 . . . #

1.150E00 2.000E-03 . . . #

1.200E00 2.000E-03 . . . #

1.250E00 2.000E-03 . . . #

1.300E00 2.000E-03 . . . #

1.350E00 2.000E-03 . . . #

1.$00E00 2.000E-03 . . . #

1.$50E00 2.000E-03 . . . #

1.500E00 2.000E-03 . . . #

1.550E00 2.000E-03 . . . #

1.600E00 2.000E-03 . . . #

1.650E00 2.000E-03 . . . #

1.%00E00 2.000E-03 . . . #1.%50E00 2.000E-03 . . . #


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1."00E01 2.000E-03 . . . #

2.000E01 2.000E-03 . . . #

2.200E01 2.000E-03 . . . #

2.$00E01 2.000E-03 . . . #

2.600E01 2.000E-03 . . . #

2."00E01 2.000E-03 . . . #

3.000E01 2.000E-03 . . . #3.200E01 2.000E-03 . . . #

3.$00E01 2.000E-03 . . . #

3.600E01 2.000E-03 . . . #

3."00E01 2.000E-03 . . . #

$.000E01 2.000E-03 . . . #

$.200E01 2.000E-03 . . . #

$.$00E01 2.000E-03 . . . #

$.600E01 2.000E-03 . . . #

$."00E01 2.000E-03 . . . #

5.000E01 2.000E-03 . . . #

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2ame resultD The collector current holds absolutely steady at > m0 despite the factthat the battery v1 voltage varies all the way from @ to B@ volts. &t would appear

from our simulation that collector-to-emitter voltage has little effect over collectorcurrent, e!cept at very low levels /ust above @ volts. The transistor is acting as a

current regulator, allowing e!actly > m0 through the collector and no more.

ow let(s see what happens if we increase the controlling &1 current from >@ H0 to

IB H0, once again sweeping the battery G1 voltage from @ to B@ volts andgraphing the collector current)

bipolar transistor simulation

i1 0 1 dc %5u

q1 2 1 0 mod1

vammeter 3 2 dc 0

v1 3 0 dc

.model mod1 npn

.dc v1 0 50 2

.plot dc i(vammeter)

.end

type npn

is 1.00E-16

bf 100.000

nf 1.000

br 1.000

nr 1.000


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ach curve on the graph reflects the collector current of the transistor, plotted over

a range of collector-to-emitter voltages, for a given amount of base current. 2ince atransistor tends to act as a current regulator, limiting collector current to a

proportion set by the base current, it is useful to e!press this proportion as astandard transistor performance measure. 2pecifically, the ratio of collector current

to base current is nown as theBetaratio symboli'ed by the :ree letter J)

2ometimes the J ratio is designated as *hfe,* a label used in a branch ofmathematical semiconductor analysis nown as *hybrid parameters* which strives

to achieve very precise predictions of transistor performance with detailed

equations. %ybrid parameter variables are many, but they are all labeled with thegeneral letter *h* and a specific subscript. The variable *h fe* is /ust anotherstandardi'ed way of e!pressing the ratio of collector current to base current, and

is interchangeable with *J.* 7ie all ratios, J is unitless.

J for any transistor is determined by its design) it cannot be altered after

manufacture. %owever, there are so many physical variables impacting J that it israre to have two transistors of the same design e!actly match. &f a circuit design

relies on equal J ratios between multiple transistors, *matched sets* of transistorsmay be purchased at e!tra cost. %owever, it is generally considered bad design

practice to engineer circuits with such dependencies.


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wouldn(t regulatecollector current at all lie the characteristic curves show. &nstead

of the collector current curves flattening out after their brief rise as the collector-emitter voltage increases, the collector current would be directly proportional to

collector-emitter voltage, rising steadily in a straight line on the graph.

0 better transistor model, often seen in more advanced te!tboos, is this)

&t casts the transistor as a combination of diode and current source, the output of thecurrent source being set at a multiple J ratio of the base current. This model is far

more accurate in depicting the true inputEoutput characteristics of a transistor) basecurrent establishes a certain amount of collector current, rather than a certain

amount of collector-emitter resistanceas the first model implies. 0lso, this model is

favored when performing networ analysis on transistor circuits, the current sourcebeing a well-understood theoretical component. "nfortunately, using a currentsource to model the transistor(s current-controlling behavior can be misleading) in

no way will the transistor ever act as asourceof electrical energy, which the currentsource symbol implies is a possibility.

#y own personal suggestion for a transistor model substitutes a constant-currentdiode for the current source)


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2ince no diode ever acts as asourceof electrical energy, this analogy escapes the

false implication of the current source model as a source of power, while depicting

the transistor(s constant-current behavior better than the rheostat model. 0notherway to describe the constant-current diode(s action would be to refer to it asa current regulator, so this transistor illustration of mine might also be described as

adiode-current regulatormodel. The greatest disadvantage & see to this model is therelative obscurity of constant-current diodes. #any people may be unfamiliar with

their symbology or even of their e!istence, unlie either rheostats or currentsources, which are commonly nown.

REVIE$%

0 transistor is said to be in its activemode if it is operating somewherebetween fully on saturated and fully off cutoff.

ase current tends to regulate collector current. y regulate, we mean that

no more collector current may e!ist than what is allowed by the base current.

The ratio between collector current and base current is called *eta* J or

*hfe*.

J ratios are different for every transistor, and they tend to change fordifferent operating conditions.


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The common-emitter ampli(ier

0t the beginning of this chapter we saw how transistors could be used as switches,

operating in either their *saturation* or *cutoff* modes. &n the last section we sawhow transistors behave within their *active* modes, between the far limits of

saturation and cutoff. ecause transistors are able to control current in an analoginfinitely divisible fashion, they find use as amplifiers for analog signals.

;ne of the simpler transistor amplifier circuits to study is the one used previously

for illustrating the transistor(s switching ability)

&t is called the common-emitterconfiguration because ignoring the power supply

battery both the signal source and the load share the emitter lead as a commonconnection point. This is not the only way in which a transistor may be used as an

amplifier, as we will see in later sections of this chapter)

efore, this circuit was shown to illustrate how a relatively small current from a

solar cell could be used to saturate a transistor, resulting in the illumination of alamp. 6nowing now that transistors are able to *throttle* their collector currents

according to the amount of base current supplied by an input signal source, weshould be able to see that the brightness of the lamp in this circuit is controllable by

the solar cell(s light e!posure. hen there is /ust a little light shone on the solarcell, the lamp will glow dimly. The lamp(s brightness will steadily increase as morelight falls on the solar cell.


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leend

# v(1)

i(v1)

v(1)

(#)--- -2.000E00 -1.000E00 0.000E00 1.000E00 2.000E00

()--- -".000E-02 -6.000E-02 -$.000E-02 -2.000E-02 0.000E00

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

0.000E00 . . # .

3.%25E-01 . . . # .

%.1!5E-01 . . . # . .

1.02$E00 . . . # .

1.26$E00 . . . # .

1.$20E00 . . . . # .

1.$!3E00 . . . . # .

1.$%0E00 . . . . # .

1.351E00 . . . . # .

1.15$E00 . . . . # .

".%!1E-01 . . . # . .

5.$!"E-01 . . . # .

1."%%E-01 . . . # .

-1."%2E-01 . . # . .

-5.501E-01 . . # . .

-"."15E-01 . . # . .

-1.151E00 . # . . .

-1.352E00 . # . . .

-1.$%2E00 . # . . .

-1.$!1E00 . # . . .

-1.$22E00 . # . . .

-1.265E00 . # . . .

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3.0$0E-06 . . # .

3.%2$E-01 . . . # .

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1.022E00 . . . # .

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5.501E-01 . . . # .

1.""0E-01 . . . # .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The simulation plots both the input voltage an 05 signal of 1.B volt peaamplitude and >@@@ %' frequency and the current through the 1B volt battery,

which is the same as the current through the speaer. hat we see here is a full 05sine wave alternating in both positive and negative directions, and a half-wave


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The only way we can get the transistor to reproduce the entire waveform as current

through the speaer is to eep the transistor in its active mode the entire time. Thismeans we must maintain current through the base during the entire input waveform

cycle. 5onsequently, the base-emitter diode /unction must be ept forward-biased atall times. 3ortunately, this can be accomplished with the aid of a $5 bias

voltageadded to the input signal. y connecting a sufficient $5 voltage in serieswith the 05 signal source, forward-bias can be maintained at all points throughout

the wave cycle)

common-emitter amplifier

vinput 1 5 sin (0 1.5 2000 0 0)

vbias 5 0 dc 2.3

r1 1 2 1'

q1 3 2 0 mod1

rsp'r 3 $ "

v1 $ 0 dc 15

.model mod1 npn

.tran 0.02m 0.%"m

.plot tran v(1&0) i(v1)

.end

leend

# v(1)

i(v1)

v(1)

(#)--- 0.000E00 1.000E00 2.000E00 3.000E00 $.000E00

()--- -3.000E-01 -2.000E-01 -1.000E-01 0.000E00 1.000E-01

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2.300E00 . . . # . .

2.6%3E00 . . . # . .

3.020E00 . . . # .

3.322E00 . . . . # .

3.563E00 . . . . # .

3.%23E00 . . . . # .


33/99

3.%!0E00 . . . . #.

3.%6%E00 . . . . #.

3.65%E00 . . . . # .

3.$52E00 . . . . # .

3.1%%E00 . . . . # .

2."50E00 . . . # . .

2.$""E00 . . . # . .2.113E00 . . . # . .

1.%50E00 . . # . . .

1.$1!E00 . . # . . .

1.1$"E00 . . # . . .

!.$!3E-01 . #. . . .

".311E-01 . # . . .

".050E-01 . # . . .

".%!%E-01 . # . . . .

1.03!E00 . .# . . .

1.2%5E00 . . # . . .

1.5%!E00 . . # . . .

1.!2!E00 . . # . .

2.300E00 . . . # . .

2.6%3E00 . . . # . .

3.01!E00 . . . # .

3.322E00 . . . . # .

3.56$E00 . . . . # .

3.%22E00 . . . . # .

3.%!0E00 . . . . #.

3.%6"E00 . . . . #.

3.65%E00 . . . . # .

3.$51E00 . . . . # .

3.1%"E00 . . . . # .

2."51E00 . . . # . .

2.$""E00 . . . # . .

2.113E00 . . . # . .

1.%$"E00 . . # . . .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

ith the bias voltage source of >.? volts in place, the transistor remains in its active

mode throughout the entire cycle of the wave, faithfully reproducing the waveformat the speaer. otice that the input voltage measured between nodes 1 and @

fluctuates between about @.8 volts and ?.8 volts, a pea-to-pea voltage of ? volts

/ust as e!pected source voltage C 1.B volts pea. The output speaer currentvaries between 'ero and almost ?@@ m0, 18@oout of phase with the inputmicrophone signal.

The following illustration is another view of the same circuit, this time with a fewoscilloscopes *scopemeters* connected at crucial points to display all the

pertinent signals)


34/99

The need for biasing a transistor amplifier circuit to obtain full waveform

reproduction is an important consideration. 0 separate section of this chapter willbe devoted entirely to the sub/ect biasing and biasing techniques. 3or now, it is

enough to understand that biasing may be necessary for proper voltage and currentoutput from the amplifier.

ow that we have a functioning amplifier circuit, we can investigate its voltage,current, and power gains. The generic transistor used in these 2+&5 analyses has a

J of 1@@, as indicated by the short transistor statistics printout included in the te!toutput these statistics were cut from the last two analyses for brevity(s sae)

type npn

is 1.00E-16

bf 100.000

nf 1.000

br 1.000

nr 1.000

J is listed under the abbreviation *bf,* which actually stands for **eta, (orward*. &f

we wanted to insert our own J ratio for an analysis, we could have done so onthe .modelline of the 2+&5 netlist.

2ince J is the ratio of collector current to base current, and we have our load

connected in series with the collector terminal of the transistor and our source


35/99

connected in series with the base, the ratio of output current to input current is

equal to beta. Thus, our current gain for this e!ample amplifier is 1@@, or 4@ d.

Goltage gain is a little more complicated to figure than current gain for this circuit.

0s always, voltage gain is defined as the ratio of output voltage divided by inputvoltage. &n order to e!perimentally determine this, we need to modify our last

2+&5 analysis to plot output voltage rather than output current so we have twovoltage plots to compare)


vinput 1 5 sin (0 1.5 2000 0 0)

vbias 5 0 dc 2.3

r1 1 2 1'

q1 3 2 0 mod1

rsp'r 3 $ "

v1 $ 0 dc 15

.model mod1 npn

.tran 0.02m 0.%"m

.plot tran v(1&0) v($&3)

.end

leend

# v(1)

v($&3)v(1)

(#)- 0.000E00 1.000E00 2.000E00 3.000E00 $.000E00

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2.300E00 . . . # . .

2.6%3E00 . . . # . .

3.020E00 . . . # .

3.322E00 . . . . # .

3.563E00 . . . . # .

3.%23E00 . . . . # .

3.%!0E00 . . . . # .

3.%6%E00 . . . . # .

3.65%E00 . . . . # .

3.$52E00 . . . # .3.1%%E00 . . . . # .

2."50E00 . . . # . .

2.$""E00 . . . # . .

2.113E00 . . # . .

1.%50E00 . . # . . .

1.$1!E00 . . # . . .

1.1$"E00 . . # . . .

!.$!3E-01 . #. . . .

".311E-01 # . . . .

".050E-01 # . . . .

".%!%E-01 . # . . . .

1.03!E00 . .# . . .

1.2%5E00 . . # . . .1.5%!E00 . . # . . .


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1.!2!E00 . . #. . .

2.300E00 . . . # . .

2.6%3E00 . . . # . .

3.01!E00 . . . # .

3.322E00 . . . . # .

3.56$E00 . . . . # .

3.%22E00 . . . . # .3.%!0E00 . . . . # .

3.%6"E00 . . . . # .

3.65%E00 . . . . # .

3.$51E00 . . . # .

3.1%"E00 . . . . # .

2."51E00 . . . # . .

2.$""E00 . . . # . .

2.113E00 . . # . .

1.%$"E00 . . # . . .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

+lotted on the same scale from @ to 4 volts, we see that the output waveform *=*

has a smaller pea-to-pea amplitude than the input waveform *K*, in addition tobeing at a lower bias voltage, not elevated up from @ volts lie the input. 2incevoltage gain for an 05 amplifier is defined by the ratio of 05 amplitudes, we can

ignore any $5 bias separating the two waveforms. ven so, the input waveform isstill larger than the output, which tells us that the voltage gain is less than 1 a

negative d figure.

To be honest, this low voltage gain is not characteristic to allcommon-emitter

amplifiers. &n this case it is a consequence of the great disparity between the input

and load resistances. ;ur input resistance 1 here is 1@@@ L, while the loadspeaer is only 8 L. ecause the current gain of this amplifier is determinedsolely by the J of the transistor, and because that J figure is fi!ed, the current gain

for this amplifier won(t change with variations in either of these resistances.%owever, voltage gain isdependent on these resistances. &f we alter the load

resistance, maing it a larger value, it will drop a proportionately greater voltagefor its range of load currents, resulting in a larger output waveform. 7et(s tryanother simulation, only this time with a ?@ L load instead of an 8 L load)


vinput 1 5 sin (0 1.5 2000 0 0)

vbias 5 0 dc 2.3

r1 1 2 1'

q1 3 2 0 mod1

rsp'r 3 $ 30

v1 $ 0 dc 15

.model mod1 npn

.tran 0.02m 0.%"m

.plot tran v(1&0) v($&3)

.end


37/99


38/99

scale of -B to 1B volts. 7ooing closely, we can see that the output waveform *=*

crests between @ and about 9 volts) appro!imately ? times the amplitude of theinput voltage.

e can perform another computer analysis of this circuit, only this time instructing2+&5 to analy'e it from an 05 point of view, giving us pea voltage figures for

input and output instead of a time-based plot of the waveforms)


vinput 1 5 ac 1.5

vbias 5 0 dc 2.3

r1 1 2 1'

q1 3 2 0 mod1

rsp'r 3 $ 30v1 $ 0 dc 15

.model mod1 npn

.ac lin 1 2000 2000

.print ac v(1&0) v($&3)

.end

freq v(1) v($&3)

2.000E03 1.500E00 $.$1"E00

+ea voltage measurements of input and output show an input of 1.B volts and anoutput of 4.418 volts. This gives us a voltage gain ratio of >.94B? 4.418 G E 1.B G,

or 9.?8>I d.


39/99


40/99

leend

# v(1)

v(3)

v(1)

(#)-- 0.000E00 1.000E00 2.000E00 3.000E00 $.000E00

()-- 0.000E00 5.000E00 1.000E01 1.500E01 2.000E01

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2.300E00 . . . # . .

2.6%3E00 . . . # . .

3.020E00 . . . # .

3.32$E00 . . . . # .

3.56$E00 . . . . # .

3.%20E00 . . . . # .

3.%!3E00 . . . . # .

3.%%0E00 . . . . # .

3.651E00 . . . . # .

3.$5$E00 . . . . # .

3.1%!E00 . . . . # .

2."50E00 . . . # . .

2.$""E00 . . . # . .

2.113E00 . . . # . .

1.%50E00 . . # . . .

1.$1"E00 . . # . . .

1.1$!E00 . . # . . .

!.$%%E-01 . #. . . .

".2%%E-01 . # . . .

".0!1E-01 . # . . .

".%"1E-01 . # . . . .

1.035E00 . # . . .

1.2%"E00 . . # . . .

1.5%!E00 . . # . . .

1.!2"E00 . . #. . .

2.300E00 . . . # . .

2.6%2E00 . . . # . .

3.020E00 . . . # .

3.322E00 . . . . # .

3.565E00 . . . . # .

3.%22E00 . . . . # .

3.%!1E00 . . . . # .

3.%%3E00 . . . . # .

3.652E00 . . . . # .

3.$51E00 . . . . # .

3.1"1E00 . . . . # .

2."50E00 . . . # . .

2.$""E00 . . . # . .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


vinput 1 5 ac 1.5

vbias 5 0 dc 2.3

r1 1 2 1'

q1 3 2 0 mod1

rsp'r 3 $ 30v1 $ 0 dc 15


41/99

.model mod1 npn

.ac lin 1 2000 2000

.print ac v(1&0) v(3&0)

.end

freq v(1) v(3)

2.000E03 1.500E00 $.$1"E00

e still have a pea output voltage of 4.418 volts with a pea input voltage of 1.Bvolts. The only difference from the last set of simulations is thephaseof the output

voltage.

2o far, the e!ample circuits shown in this section have all used + transistors.

++ transistors are /ust as valid to use as + in anyamplifier configuration, solong as the proper polarity and current directions are maintained, and the common-

emitter amplifier is no e!ception. The inverting behavior and gain properties of a++ transistor amplifier are the same as its + counterpart, /ust the polarities are

different)

REVIE$%

Common-emittertransistor amplifiers are so-called because the input and

output voltage points share the emitter lead of the transistor in common with

each other, not considering any power supplies.


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1.03!E00 . .# . . .

1.2%5E00 . . # . . .

1.5%!E00 . . # . . .

1.!2!E00 . . #. . .

2.300E00 . . . # . .

2.6%3E00 . . . # . .

3.01!E00 . . . # .3.322E00 . . . . # .

3.56$E00 . . . . # .

3.%22E00 . . . . # .

3.%!0E00 . . . # .

3.%6"E00 . . . # .

3.65%E00 . . . . # .

3.$51E00 . . . . # .

3.1%"E00 . . . . # .

2."51E00 . . . # . .

2.$""E00 . . . # . .

2.113E00 . . . # . .

1.%$"E00 . # . . .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

%ere(s another view of the circuit, this time with oscilloscopes connected to several

points of interest)

2ince this amplifier configuration doesn(t provide any voltage gain in fact, inpractice it actually has a voltage gain of slightly lessthan 1, its only amplifying

factor is current. The common-emitter amplifier configuration e!amined in theprevious section had a current gain equal to the J of the transistor, being that the

input current went through the base and the output load current went through thecollector, and J by definition is the ratio between the collector and emitter currents.

&n the common-collector configuration, though, the load is situated in series withthe emitter, and thus its current is equal to the emitter current. ith the emitter


48/99


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52/99

The final transistor amplifier configuration we need to study is the common-base.

This configuration is more comple! than the other two, and is less common due toits strange operating characteristics.

&t is called the common-baseconfiguration because $5 power source aside, the

signal source and the load share the base of the transistor as a common connectionpoint)

+erhaps the most striing characteristic of this configuration is that the input signalsource must carry the full emitter current of the transistor, as indicated by the heavyarrows in the first illustration. 0s we now, the emitter current is greater than any

other current in the transistor, being the sum of base and collector currents. &n thelast two amplifier configurations, the signal source was connected to the base lead

of the transistor, thus handling the leastcurrent possible.

ecause the input current e!ceeds all other currents in the circuit, including the

output current, the current gain of this amplifier is actually less than notice howloadis connected to the collector, thus carrying slightly less current than the signal


53/99

source. &n other words, it attenuatescurrent rather than amplifyingit. ith

common-emitter and common-collector amplifier configurations, the transistorparameter most closely associated with gain was J. &n the common-base circuit, we

follow another basic transistor parameter) the ratio between collector current andemitter current, which is a fraction always less than 1. This fractional value for any

transistor is called the alpharatio, or N ratio.

2ince it obviously can(t boost signal current, it only seems reasonable to e!pect it to

boost signal voltage. 0 2+&5 simulation will vindicate that assumption)

common-base amplifier

vin 0 1

r1 1 2 100

q1 $ 0 2 mod1

v1 3 0 dc 15

rload 3 $ 5'

.model mod1 npn

.dc vin 0.6 1.2 .02

.plot dc v(3&$)

.end

v(3&$) 0.000E00 5.000E00 1.000E01 1.500E01 2.000E01

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

5.!13E-03 # . . . .

1.2%$E-02 # . . . .

2.%30E-02 # . . . .

5.%%6E-02 # . . . .

1.1!3E-01 # . . . .

2.35"E-01 .# . . . .

$.3%0E-01 .# . . . .

%.$$%E-01 . # . . . .

1.163E00 . # . . . .1.6"2E00 . # . . . .

2.2"1E00 . # . . . .


54/99


55/99


56/99

-1.502E-02 . . # . . .

1.$!6E-02 . . . # . .

$.$00E-02 . . . # . .

%.0$"E-02 . . # . .

!.21$E-02 . . . #. .

1.0"1E-01 . . . .# .

1.1%5E-01 . . . . # .1.1!6E-01 . . . . # .

1.136E-01 . . . . # .

1.00!E-01 . . . # .

".203E-02 . . . # . .

5.%6$E-02 . . . # . .

2.!%0E-02 . . . # . .

-1.$$0E-05 . . # . .

-2.!"1E-02 . . # . . .

-5.%55E-02 . . # . . .

-".1%"E-02 . . # . . .

-1.011E-01 . # . . .

-1.13"E-01 . # . . . .

-1.1!2E-01 . # . . . .

-1.1%$E-01 . # . . . .

-1.0"5E-01 . #. . . .

-!.20!E-02 . .# . . .

-%.020E-02 . . # . . .

-$.$0%E-02 . # . . .

-1.502E-02 . . # . . .

1.$!6E-02 . . . # . .

$.$1%E-02 . . . # . .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

0s you can see, the input and output waveforms are in phase with each other. Thistells us that the common-base amplifier is non-inverting.

common-base amplifier

vin 0 1 ac 0.12

vbias 1 5 dc 0.!5

r1 5 2 100

q1 $ 0 2 mod1

v1 3 0 dc 15

rload 3 $ 5'

.model mod1 npn

.ac lin 1 2000 2000

.print ac v(1&0) v(3&$)

.end

freq v(1) v(3&$)

2.000E03 1.200E-01 5.12!E00


57/99

Goltage figures from the second analysis 05 mode show a voltage gain of 4>.I4>

B.1>9 G E @.1> G, or ?>.A1I d)

%ere(s another view of the circuit, showing the phase relations and $5 offsets ofvarious signals in the circuit /ust simulated)

. . . and for a ++ transistor)


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common-base amplifier +, bias 0."5 volts

vin 0 1 ac 0.0"

vbias 1 5 dc 0."5

r1 5 2 100

q1 $ 0 2 mod1

v1 3 0 dc 15

rload 3 $ 5'

.model mod1 npn

.ac lin 1 2000 2000

.print ac v(1&0) v(3&$)

.end

freq v(1) v(3&$)

2.000E03 ".000E-02 3.005E00

common-base amplifier dc bias 0.! volts

vin 0 1 ac 0.0"

vbias 1 5 dc 0.!0

r1 5 2 100

q1 $ 0 2 mod1

v1 3 0 dc 15rload 3 $ 5'

.model mod1 npn

.ac lin 1 2000 2000

.print ac v(1&0) v(3&$)

.end

freq v(1) v(3&$)

2.000E03 ".000E-02 3.26$E00

common-base amplifier dc bias 0.!5 volts

vin 0 1 ac 0.0"

vbias 1 5 dc 0.!5

r1 5 2 100

q1 $ 0 2 mod1

v1 3 0 dc 15

rload 3 $ 5'

.model mod1 npn

.ac lin 1 2000 2000

.print ac v(1&0) v(3&$)


60/99

.end

freq v(1) v(3&$)

2.000E03 ".000E-02 3.$1!E00

0 trend should be evident here) with increases in $5 bias voltage, voltage gain

increases as well. e can see that the voltage gain is increasing because eachsubsequent simulation produces greater output voltage for the e!act same input

signal voltage @.@8 volts. 0s you can see, the changes are quite large, and they arecaused by miniscule variations in bias voltageD

The combination of very low current gain always less than 1 and somewhatunpredictable voltage gain conspire against the common-base design, relegating it

to few practical applications.

REVIE$%

Common-basetransistor amplifiers are so-called because the input and

output voltage points share the base lead of the transistor in common with

each other, not considering any power supplies.

The current gain of a common-base amplifier is always less than 1. The

voltage gain is a function of input and output resistances, and also theinternal resistance of the emitter-base /unction, which is sub/ect to change

with variations in $5 bias voltage. 2uffice to say that the voltage gain of acommon-base amplifier can be very high.

The ratio of a transistor(s collector current to emitter current is called N. The

N value for any transistor is always less than unity, or in other words, less

than 1.

Biasin+ techni,ues

&n the common-emitter section of this chapter, we saw a 2+&5 analysis where the

output waveform resembled a half-wave rectified shape) only half of the inputwaveform was reproduced, with the other half being completely cut off. 2ince our

purpose at that time was to reproduce the entire waveshape, this constituted a

problem. The solution to this problem was to add a small bias voltage to the


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waveform cycle. Together, however, they function as a team to produce an output

waveform identical in shape to the input waveform.

0 decided advantage of the class push-pull amplifier design over the class 0

design is greater output power capability. ith a class 0 design, the transistordissipates a lot of energy in the form of heat because it never stops conducting

current. 0t all points in the wave cycle it is in the active conducting mode,conducting substantial current and dropping substantial voltage. This means there is

substantial power dissipated by the transistor throughout the cycle. &n a class design, each transistor spends half the time in cutoff mode, where it dissipates 'ero

power 'ero current C 'ero power dissipation. This gives each transistor a time to*rest* and cool while the other transistor carries the burden of the load. 5lass 0

amplifiers are simpler in design, but tend to be limited to low-power signalapplications for the simple reason of transistor heat dissipation.

There is another class of amplifier operation nown as class AB, which issomewhere between class 0 and class ) the transistor spends more than B@O but

less than 1@@O of the time conducting current.

&f the input signal bias for an amplifier is slightly negative opposite of the bias

polarity for class 0 operation, the output waveform will be further *clipped* than itwas with class biasing, resulting in an operation where the transistor spends the

ma/ority of the time in cutoff mode)

0t first, this scheme may seem utterly pointless. 0fter all, how useful could anamplifier be if it clips the waveform as badly as this< &f the output is used directly

with no conditioning of any ind, it would indeed be of questionable utility.

%owever, with the application of a tan circuit parallel resonant inductor-capacitorcombination to the output, the occasional output surge produced by the amplifier


64/99

can set in motion a higher-frequency oscillation maintained by the tan circuit. This

may be liened to a machine where a heavy flywheel is given an occasional *ic*to eep it spinning)

5alled class Coperation, this scheme also en/oys high power efficiency due to the

fact that the transistors spend the vast ma/ority of time in the cutoff mode, wherethey dissipate 'ero power. The rate of output waveform decay decreasing

oscillation amplitude between *ics* from the amplifier is e!aggerated here for

the benefit of illustration. ecause of the tuned tan circuit on the output, this typeof circuit is usable only for amplifying signals of definite, fi!ed frequency.

0nother type of amplifier operation, significantly different from 5lass 0, , 0, or

5, is called Class . &t is not obtained by applying a specific measure of biasvoltage as are the other classes of operation, but requires a radical re-design of the

amplifier circuit itself. &t(s a little too early in this chapter to investigate e!actly howa class $ amplifier is built, but not too early to discuss its basic principle of

operation.

0 class $ amplifier reproduces the profile of the input voltage waveform bygenerating a rapidly-pulsing squarewave output. The duty cycle of this outputwaveform time *on* versus total cycle time varies with the instantaneous

amplitude of the input signal. The following plots demonstrate this principle)


65/99

The greater the instantaneous voltage of the input signal, the greater the duty cycle

of the output squarewave pulse. &f there can be any goal stated of the class $design, it is to avoid active-mode transistor operation. 2ince the output transistor of

a class $ amplifier is never in the active mode, only cutoff or saturated, there willbe little heat energy dissipated by it. This results in very high power efficiency for

the amplifier. ;f course, the disadvantage of this strategy is the overwhelmingpresence of harmonics on the output. 3ortunately, since these harmonic frequencies

are typically much greater than the frequency of the input signal, they can befiltered out by a low-pass filter with relative ease, resulting in an output more

closely resembling the original input signal waveform. 5lass $ technology istypically seen where e!tremely high power levels and relatively low frequenciesare encountered, such as in industrial inverters devices converting $5 into 05

power to run motors and other large devices and high-performance audioamplifiers.

0 term you will liely come across in your studies of electronics is somethingcalled "uiescent, which is a modifier designating the normal, or 'ero input signal,condition of a circuit. Fuiescent current, for e!ample, is the amount of current in a

circuit with 'ero input signal voltage applied. ias voltage in a transistor circuitforces the transistor to operate at a different level of collector current with 'eroinput signal voltage than it would without that bias voltage. Therefore, the amount

of bias in an amplifier circuit determines its quiescent values.

&n a class 0 amplifier, the quiescent current should be e!actly half of its saturationvalue halfway between saturation and cutoff, cutoff by definition being 'ero.

5lass and class 5 amplifiers have quiescent current values of 'ero, since they are


66/99

supposed to be cutoff with no signal applied. 5lass 0 amplifiers have very low

quiescent current values, /ust above cutoff. To illustrate this graphically, a *loadline* is sometimes plotted over a transistor(s characteristic curves to illustrate its

range of operation while connected to a load resistance of specific value)

0 load line is a plot of collector-to-emitter voltage over a range of base currents. 0t

the lower-right corner of the load line, voltage is at ma!imum and current is at 'ero,representing a condition of cutoff. 0t the upper-left corner of the line, voltage is at

'ero while current is at a ma!imum, representing a condition of saturation. $otsmaring where the load line intersects the various transistor curves represent

realistic operating conditions for those base currents given.

Fuiescent operating conditions may be shown on this type of graph in the form of a

single dot along the load line. 3or a class 0 amplifier, the quiescent point will be inthe middle of the load line, lie this)


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ow that we now a little more about the consequences of different $5 bias

voltage levels, it is time to investigate practical biasing techniques. 2o far, &(veshown a small $5 voltage source battery connected in series with the 05 input

signal to bias the amplifier for whatever desired class of operation. &n real life, theconnection of a precisely-calibrated battery to the input of an amplifier is simply

not practical. ven if it were possible to customi'e a battery to produce /ust the

right amount of voltage for any given bias requirement, that battery would notremain at its manufactured voltage indefinitely. ;nce it started to discharge and itsoutput voltage drooped, the amplifier would begin to drift in the direction of class

operation.

Tae this circuit, illustrated in the common-emitter section for a 2+&5 simulation,

for instance)


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5ombining these two separate analyses, we get a superposition of almost 1.B volts

05 and >.? volts $5, ready to be connected to the base of the transistor)

nough tal -- it(s about time for a 2+&5 simulation of the whole amplifier circuit.

&(ll use a capacitor value of 1@@ H3 to obtain an arbitrarily low @.I9A L impedanceat >@@@ %')

voltae divider biasin

vinput 1 0 sin (0 1.5 2000 0 0)

c1 1 5 100u

r1 5 2 1'r2 $ 5 "$66

r3 5 0 1533

q1 3 2 0 mod1

rsp'r 3 $ "

v1 $ 0 dc 15

.model mod1 npn

.tran 0.02m 0.%"m


.end

leend

# v(1)

i(v1)

v(1)

(#)-- -2.000E00 -1.000E00 0.000E00 1.000E00 2.000E00

()-- -3.000E-01 -2.000E-01 -1.000E-01 0.000E00 1.000E-01

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

0.000E00 . . # . .

3.%30E-01 . . # . .

%.1!%E-01 . . . # . .

1.022E00 . . . # .

1.263E00 . . . . # .

1.$23E00 . . . # .1.$!0E00 . . . . # .

1.$6%E00 . . . . # .


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0 voltage divider(s output depends not only on the si'e of its constituent resistors,

but also on how much current is being divided away from it through a load. &n thiscase, the base-emitter + /unction of the transistor is a load that decreases the $5

voltage dropped across ?, due to the fact that the bias current /oins with ?(scurrent to go through >, upsetting the divider ratio formerly set by the resistance

values of >and ?. &n order to obtain a $5 bias voltage of >.? volts, the values of>andEor ?must be ad/usted to compensate for the effect of base current loading.

&n this case, we want to increasethe $5 voltage dropped across ?, so we canlower the value of >, raise the value of ?, or both.

voltae divider biasin

vinput 1 0 sin (0 1.5 2000 0 0)

c1 1 5 100u

r1 5 2 1'

r2 $ 5 6' --- /2 decreased to 6 ' oms

r3 5 0 $' --- /3 increased to $ ' oms

q1 3 2 0 mod1

rsp'r 3 $ "

v1 $ 0 dc 15

.model mod1 npn

.tran 0.02m 0.%"m


.end

leend

# v(1)

i(v1)

v(1)

(#)-- -2.000E00 -1.000E00 0.000E00 1.000E00 2.000E00

()-- -3.000E-01 -2.000E-01 -1.000E-01 0.000E00 1.000E-01

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

0.000E00 . . # . .

3.%30E-01 . . . # . .

%.1!%E-01 . . . # . .

1.022E00 . . . # .


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sinusoidal signals will tend to be distorted, as the capacitor responds differently to

each of the signal(s constituent harmonics. 0n e!treme e!ample of this would be alow-frequency square-wave signal)

&ncidentally, this same problem occurs when oscilloscope inputs are set to the *05coupling* mode. &n this mode, a coupling capacitor is inserted in series with the

measured voltage signal to eliminate any vertical offset of the displayed waveformdue to $5 voltage combined with the signal. This wors fine when the 05

component of the measured signal is of a fairly high frequency, and the capacitoroffers little impedance to the signal. %owever, if the signal is of a low frequency,

andEor contains considerable levels of harmonics over a wide frequency range, theoscilloscope(s display of the waveform will not be accurate.


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This new feedbac resistor drops voltage proportional to the emitter current

through the transistor, and it does so in such a way as to oppose the input signal(sinfluence on the base-emitter /unction of the transistor. 7et(s tae a closer loo at

the emitter-base /unction and see what difference this new resistor maes)

ith no feedbac resistor connecting the emitter to ground, whatever level of input

signal Ginput maes it through the coupling capacitor and 1E>E?resistor networwill be impressed directly across the base-emitter /unction as the transistor(s input

voltage G-. &n other words, with no feedbac resistor, G-equals Ginput.Therefore, if Ginputincreases by 1@@ mG, then G-liewise increases by 1@@ mG) a

change in one is the same as a change in the other, since the two voltages are equalto each other.

ow let(s consider the effects of inserting a resistor feedbac between thetransistor(s emitter lead and ground)


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ote how the voltage dropped across feedbacadds with G-to equal Ginput. ith

feedbacin the Ginput-- G-loop, G-will no longer be equal to Ginput. e now thatfeedbacwill drop a voltage proportional to emitter current, which is in turn

controlled by the base current, which is in turn controlled by the voltage droppedacross the base-emitter /unction of the transistor G-. Thus, if Ginputwere to

increase in a positive direction, it would increase G-, causing more base current,causing more collector load current, causing more emitter current, and causing

more feedbac voltage to be dropped across feedbac. This increase of voltage dropacross the feedbac resistor, though,subtractsfrom Ginputto reduce the G-, so that

the actual voltage increase for G-will be less than the voltage increase of G input.o longer will a 1@@ mG increase in Ginputresult in a full 1@@ mG increase for G-,

because the two voltages are notequal to each other.

5onsequently, the input voltage has less control over the transistor than before, andthe voltage gain for the amplifier is reduced) /ust what we e!pected from negativefeedbac.

&n practical common-emitter circuits, negative feedbac isn(t /ust a lu!uryP it(s anecessity for stable operation. &n a perfect world, we could build and operate a

common-emitter transistor amplifier with no negative feedbac, and have the fullamplitude of Ginputimpressed across the transistor(s base-emitter /unction. This

would give us a large voltage gain. "nfortunately, though, the relationship betweenbase-emitter voltage and base-emitter current changes with temperature, as

predicted by the *diode equation.* 0s the transistor heats up, there will be less of aforward voltage drop across the base-emitter /unction for any given current. This

causes a problem for us, as the 1E>voltage divider networ is designed to providethe correct quiescent current through the base of the transistor so that it will operate

in whatever class of operation we desire in this e!ample, &(ve shown the amplifierworing in class-0 mode. &f the transistor(s voltageEcurrent relationship changes

with temperature, the amount of $5 bias voltage necessary for the desired class ofoperation will change. &n this case, a hot transistor will draw more bias current for

the same amount of bias voltage, maing it heat up even more, drawing even more

bias current. The result, if uncheced, is called thermal runaway.


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base-emitter voltage of a transistor constant, then its emitter current should liewise

be constant, given a constant temperature)

This constant emitter current, multiplied by a constant N ratio, gives a constantcollector current through load, provided that there is enough battery voltage to eep

the transistor in its active mode for any change in load(s resistance.

#aintaining a constant voltage across the transistor(s base-emitter /unction is easy)

use a forward-biased diode to establish a constant voltage of appro!imately @.Ivolts, and connect it in parallel with the base-emitter /unction)

ow, here(s where it gets interesting. The voltage dropped across the diodeprobably won(t be @.I volts e!actly. The e!act amount of forward voltage dropped

across it depends on the current through the diode, and the diode(s temperature, allin accordance with the diode equation. &f diode current is increased say, by

reducing the resistance of bias, its voltage drop will increase slightly, increasingthe voltage drop across the transistor(s base-emitter /unction, which will increase

the emitter current by the same proportion, assuming the diode(s + /unction andthe transistor(s base-emitter /unction are well-matched to each other. &n other words,

transistor emitter current will closely equal diode current at any given time. &f you

change the diode current by changing the resistance value of bias, then thetransistor(s emitter current will follow suit, because the emitter current is described


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Alpha) the ratio of collector current to emitter current, N may be derived from J,

being equal to JEJ=1.

ipolar transistors come in a wide variety of physical pacages. +acage type is

primarily dependent upon the power dissipation of the transistor, much lieresistors) the greater the ma!imum power dissipation, the larger the device has to

be to stay cool. There are several standardi'ed pacage types for three-terminalsemiconductor devices, any of which may be used to house a bipolar transistor.

This is an important fact to consider) there are many other semiconductor devicesother than bipolar transistors which have three connection points. &t is impossibleto

positively identify a three-terminal semiconductor device without referencing thepart number printed on it, andEor sub/ecting it to a set of electrical tests.

Documents

Lessons in Electric Circuits Bipolar Junction Transistor